RRB ALP CBT I 25 Nov 2024 Shift 3 Paper

Concept:The angles of depression from the tower top equal the angles of elevation from the roads. Use tangent ratios to find each horizontal distance from the tower base, then add them because the tower lies between the two parallel roads.Explanation:Let the tower be $AC$ with height $10\,m$.Let the two roads be at points $B$ and $D$ on opposite sides of the tower base $C$.So the total distance between the roads is $BD = BC + CD$.For the first road at angle of depression $30^\circ$:In $\triangle ABC$, $\tan 30^\circ = \frac{AC}{BC}$.$\Rightarrow \frac{1}{\sqrt{3}} = \frac{10}{BC}$$\Rightarrow BC = 10\sqrt{3} = 10 \times 1.732 = 17.32\,m$For the second road at angle of depression $45^\circ$:In $\triangle ACD$, $\tan 45^\circ = \frac{AC}{CD}$.$\Rightarrow 1 = \frac{10}{CD}$$\Rightarrow CD = 10\,m$Therefore, distance between the roads:$BD = BC + CD = 17.32 + 10 = 27.32\,m$Answer:The roads are approximately $27.32\,m$ apart, which matches option B.